MHT CET 2023 9th May Morning Shift | Fluid Mechanics Question 82 | Physics

The radii of two soap bubbles are $$r_1$$ and $$r_2$$. In isothermal condition they combine with each other to form a single bubble. The radius of resultant bubble is

$$\mathrm{R}=\frac{\mathrm{r}_1+\mathrm{r}_2}{2}$$

$$\mathrm{R}=\mathrm{r}_1\left(\mathrm{r}_1 \mathrm{r}_2+\mathrm{r}_2\right)$$

$$\mathrm{R}=\sqrt{\mathrm{r}_1^2+\mathrm{r}_2^2}$$

$$\mathrm{R}=\mathrm{r}_1+\mathrm{r}_2$$

MHT CET 2022 11th August Evening Shift

MCQ (Single Correct Answer)

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The work done in blowing a soap bubble of radius $$\mathrm{R}$$ is '$$\mathrm{W}_1$$' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $$2 \mathrm{R}$$ is blown and the work done is '$$\mathrm{W}_2$$'. Then

$$\mathrm{W}_2=\mathrm{W}_1$$

$$\mathrm{W}_2=4 \mathrm{~W}_1$$

$$\mathrm{W}_2<4 \mathrm{~W}_1$$

$$\mathrm{W}_2=0$$

MHT CET 2022 11th August Evening Shift

MCQ (Single Correct Answer)

-0

A steel coin of thickness '$$\mathrm{d}$$' and density '$$\rho$$' is floating on water of surface tension '$$T$$'. The radius of the coin $$(R)$$ is [$$\mathrm{g}=$$ acceleration due to gravity]

$$\frac{\mathrm{T}}{\rho \mathrm{gd}}$$

$$\frac{4 \mathrm{~T}}{3 \rho g \mathrm{~d}}$$

$$\frac{3 \mathrm{~T}}{4 \rho \mathrm{gd}}$$

$$\frac{2 \mathrm{~T}}{\rho \mathrm{gd}}$$

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